25x^2+58x+16=0

Solution for 25x^2+58x+16=0 equation:

25x^2+58x+16=0

a = 25; b = 58; c = +16;
Δ = b2-4ac
Δ = 582-4·25·16
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-42}{2*25}=\frac{-100}{50}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+42}{2*25}=\frac{-16}{50}
=-8/25
$